MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose to have a Lie algebra L with a reductive lie subalgebra G. Let l an element of L such that [l,g] is in G for every g in G, is it true that l is an element of G?if not, there are some restriction on G that makes it true?

share|cite|improve this question
The answer to your first question is rather obviously no, since if $L = G \times H$ with H abelian then every element $l\in H$ has $[l,g]=0\in G$. – Robin Chapman Apr 20 '10 at 14:26
It is true.But If L doesn't decompose like that? – Michele Torielli Apr 20 '10 at 14:33
The fact is that I have in mind a specific example: L is the set of weight zero vector fields in C^n and G is the set of vector fields tangent to an hypersurface. – Michele Torielli Apr 20 '10 at 14:39
up vote 4 down vote accepted

Sorry, this started as a comment, but got too long.

If $G$ is semisimple, then every derivation of $G$ is inner, so that the normalizer $N_L(G)=C_L(G)+G$ where $C_L(G)$ is the centralizer. In this situtaion Michele's condition holds if and only if the centralizer is trivial.

However the case where $G$ is reductive isn't so easy. A reductive $G$ is the direct sum of an Abelian and a semisimple Lie algebra. The Abelian part can have (if at least two-dimensional) non-trivial derivations. These will lift to non-inner derivations of $G$. If we let $L$ be the semidirect product of $G$ with a one-dimensional Lie algebra using this derivation, then $L$ normalizes $G$, $L\ne G$ but $C_L(G)$ is trivial.

share|cite|improve this answer

EDIT: Realized I had misread the question.

The set of $l$ such that $[l,g]\in G$ for all $G$ is called the normalizer of $G$. It's not particularly common for the normalizer to be the same as G, though it does happen sometimes. There are a few theorems I know about specific Lie algebras being self-normalizing, such as Cartan and Borel subalgebras in reductive Lie algebras, but it's not very common and I don't know any general condition which guarantees it.

share|cite|improve this answer
L/G is not necessarily a Lie algebra, right? Do you mean that the centralizer of G in L should be trivial? – damiano Apr 20 '10 at 14:59
thanks!are there restriction on L and G to make the centralizer to be zero? – Michele Torielli Apr 20 '10 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.